In a regular rectangular pyramid, the side of the base is 6, and the total surface area of the pyramid

In a regular rectangular pyramid, the side of the base is 6, and the total surface area of the pyramid is 27√3. Find the volume of the pyramid.

ΔABC – equilateral:

AC = BC = AB = 6.

Point O is the intersection of medians and heights.

SO is the height of the pyramid.

HC = AC / 2 = 3.

BH = √ (BC ^ 2 – HC ^ 2) = √ (36 – 9) = 3√3;

Base area:

SABC = (AC * BH) / 2 = (6 * 3√3) / 2 = 9√3;

At the point of intersection, the medians are divided in a ratio of 2: 1:

OH = BH / 3 = (3√3) / 3 = √3.

Side edge height HS:

HS = √ (HO ^ 2 + OS ^ 2) = √ ((√3) ^ 2 + OS ^ 2) = √ (3 + OS ^ 2);

The area of ​​one side face:

SASC = (AC * HS) / 2 = (6 * √ (3 + OS ^ 2)) / 2 = 3√ (3 + OS ^ 2);

Total surface area:

S = SABC + 3 * SASC = 9√3 + 3 * 3√ (3 + OS ^ 2);

By condition S = 27√3:

9√3 + 9√ (3 + OS ^ 2) = 27√3;

9√ (3 + OS ^ 2) = 18√3;

√ (3 + OS ^ 2) = 2√3;

3 + OS ^ 2 = 4 * 3;

OS ^ 2 = 9;

OS = 3.

Pyramid volume:

V = (SABC * OS) / 3 = (9√3 * 3) / 3 = 9√3 ≈ 15.6.

Answer: V = 9√3.